Question

所以我有这张表A：

color        ID       MODEL
-----------------------------
red        | 10   |   HONDA
blue       | 10   |   TOYOTA
red        | 15   |   ISUZU
red        | 30   |   MITSUBISHI
red        | 5    |   HONDA
blue       | 5    |   SUBARU
orange     | 10   |   HYUNDAI
black      | 40   |   CHRYSLER

我希望得到彼此不具有相同身份的所有红色和蓝色

所以我的预期结果是：

color  ID
------------
red  | 15
red  | 30

Answer 1

或使用anti join

select t1.color, t1.id 
from 
    tableA t1
left outer join 
     tableA t2 on t2.id = t1.id and t2.color != t1.color 
where 
 t1.color in ('red', 'blue')
and t2.color is null

Answer 2

像这样的东西

SELECT * 
FROM   yourtable 
WHERE  id IN (SELECT id 
              FROM   youtable 
              GROUP  BY id 
              HAVING Count(case when color IN ( 'red', 'blue' ) then 1 end) = 1)

Answer 3

使用NOT EXISTS查找ID匹配但颜色不同的行：

select *
from yourtable a
where a.color IN ('red', 'blue')
  and not exists (
    select 1
    from yourtable b
    where a.id = b.id
      and b.color NOT IN ('red', 'blue')
    )

注意：

有效查找考虑添加索引（更多不同的值意味着更高的树遍历效率）
考虑使用其他字典color表格

Answer 4

我认为你的预期结果应该是红色15.红色30不符合条件，因为黄色30，它们具有相同的ID。看看我的代码：

SELECT t.id, t.color
FROM t
INNER JOIN
(SELECT id
 FROM t
 GROUP BY id
 HAVING COUNT(*) = 1) sub
ON t.id = sub.id
WHERE color = 'red'
OR color = 'blue'

Answer 5

删除了上一个答案。经过测试，它正在运行。

WITH tbl
AS (SELECT
  color,
  id
FROM TABLE1
WHERE color IN ('red', 'blue')
)
SELECT
  t1.color,
  t1.id
FROM tbl t1
LEFT JOIN tbl t2
  ON t1.id = t2.id
  AND t1.color <> t2.color
WHERE t1.id IS NULL
OR t2.id IS NULL

Answer 6

此查询返回COLOR为红色或蓝色但不是两者的所有ID。

select id , color
from  your_table 
where color in ('red', 'blue')
and id in (
    select id 
    from   your_table 
    where color in ('red', 'blue')
    group by id 
    having count(*) = 1)

如何选择在另一列中不匹配的数据

6 个答案: